I'm going through "Probabilistic Machine Learning: An Introduction" by Murphy (2021), and currently trying to do Ex. 3.1 (p. 80), which states:

Let $X \sim N(0, 1)$ and $Y = WX$, where $p(W = 1) = p(W =-1) = .5$. It is clear that X and Y are not independent, since Y is a function of X.

a. Show $Y \sim N(0, 1)$

b. Show $Cov[X, Y] = 0$ Thus X and Y are uncorrelated but dependent, even though they are Gaussian. Hint: use the definition of covariance:

$Cov[X,Y] = \mathbb{E}[XY]−\mathbb{E}[X]\mathbb{E}[Y]$ and $\mathbb{E}[XY ] = \mathbb{E}[\mathbb{E}[XY |W]]$

a) Now, conceptually, I get that $\mathbb{E}[W] = 0$ and that $Y = WX$ should be distributed $N(0, 1)$. But this is just because $W$ is such a simple distribution, and I would definitely be stuck if it were more complicated. So everything in between is done in my head because "it seems like it should be so", and not because I can derive it.

b) I also get that $\mathbb{E}[XY]$ and $\mathbb{E}[X]\mathbb{E}[Y]$ should be 0, therefore $Cov[EX]=0$, but also only on an abstract basis.

Could someone please guide me through this? This is not a homework question or any assignment, just trying to plough through the book by myself.


  • $\begingroup$ $Y$ is a mixture of two identically distributed symmetric variables and therefore has the same distribution. When those variables have a finite variance, the covariance must be zero by an analogous symmetry argument--no formulas are needed. See "Solution by circular symmetry" at stats.stackexchange.com/a/257919/919 for how one goes about making this kind of argument. $\endgroup$
    – whuber
    Feb 23, 2021 at 13:27
  • 1
    $\begingroup$ @whuber Thanks for the explanation and the link. $\endgroup$
    – Zlo
    Feb 23, 2021 at 13:44

1 Answer 1


Slight Generalization

The distribution of $Y$ can be factorized as $$p_Y(y) = \sum_{w} p_{Y,W}(y,w) = \sum_{w}p_W(w)p_{Y\mid W}(y\mid w) $$ If you denote $P(W=1)=q$ you obtain that $Y$ is distributed as $$ p_Y(y) = q\,p_X(x) + (1-q)\,p_X(-x)$$ However in your case you have a distribution which is symmetric about zero so that $p_X(x) = p_X(-x)$, therefore you obtain $p_Y = (q + 1 - q)p_X = p_X$.

Make it more general

You could make the problem more complex by having $W$ take $K$ non-zero values and use the same formulas. However you will need to derive the pdf of $Y\mid W=w$ which is equal to the pdf of $wX$: this is simply $\frac{1}{\lvert w \rvert}p_X(y/w)$.

Therefore you have that $Y$ is distributed as $$ p_Y(y) = \sum_{i=1}^K \dfrac{p(w_i)}{\lvert w_i \rvert}p_X(y/w_i)$$

  • $\begingroup$ Thanks, very helpful! $\endgroup$
    – Zlo
    Feb 23, 2021 at 11:34

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